Question: $f(x, y) = \left( x^2, y^2 \right)$ What is the divergence of $f$ at $(-2, 1)$ ?
Answer: The formula for divergence in two dimensions is $\text{div}(f) = \dfrac{\partial P}{\partial x} + \dfrac{\partial Q}{\partial y}$, where $P$ is the $x$ -component of $f$ and $Q$ is the $y$ -component. Let's differentiate! $\begin{aligned} \dfrac{\partial P}{\partial x} &= \dfrac{\partial}{\partial x} \left[ x^2 \right] \\ \\ &= 2x \\ \\ \dfrac{\partial Q}{\partial y} &= \dfrac{\partial}{\partial y} \left[ y^2 \right] \\ \\ &= 2y \end{aligned}$ Adding the two partial derivatives, $\text{div}(f) = 2x + 2y$. The divergence of $f$ at $(-2, 1)$ is $-2$.